We want to evaluate the given line integral
along the curve C,
where C is the arc of the curve Y equals E
raised to the power of X
for X greater than or equal to zero,
and less than or equal to three.
You'll notice how in this example,
the line integral is written in differential form,
so our process will be a little bit different.
We want to begin by parameterizing the curve C,
meaning we need to write parametric equations
that will trace out the curve C.
So if we have Y equals E to the X over this interval,
the easiest parameterization is probably going to be
let X equal T or X of T
be equal to T.
So if X is equal to T,
we can write Y or Y of T
as Y of T equals
E raised to the power of T,
and because we have X equals T,
the interval for T is also going to be
a closed interval from zero to three.
And we're also going to need X prime of T
and Y prime of T.
Let's go ahead and find those now.
X prime of T is equal to the derivative of T
with respect to T, which is one,
and Y prime of T
is equal to the derivative of E to the T,
which is just E to the T.
Now looking at our notes below
because the given line integral
is in differential form,
or this form here.
For the next step, let's write this as
two separate line integrals
in this form here.
So the given line integral is equal to the line integral
along the curve C of X,
integrated with respect to Y,
plus the line integral along the curve C of Y,
integrated with respect to X.
Now for our next step,
looking at these notes here,
we'll write the integrated function,
or the function of T using the parameteric equations,
and the differential X is equal to
X prime of T, DT,
and differential Y is equal to
Y prime of T, DT.
So for the first line integral,
we'll have the integral of,
well, X is equal to T,
so we'll have T,
and again, differential Y is equal to
Y prime of T, DT,
which would be E to the T, DT.
Limits of integration for T are going to be
from zero to three,
and we'll have plus the integral
of Y, which is equal to E to the T,
and then differential X is equal to
X prime of T,
which is equal to one times DT.
So we just have DT.
The limits of integration for T, again,
are from zero to three.
Now let's evaluate these on the next slide.
In order to evaluate this line integral,
we'll have to perform integration by parts,
and I've included the formula here below for review.
So applying integration by parts,
let's let U be equal to T,
and therefore differential U
is equal to one times DT, or just DT.
So if we let U equal T,
then E to the T, DT would be equal to DV.
So if DV is equal to E to the T, DT,
we need to integrate both sides of the equation
to find V.
Well, the integral of E to the T with respect to T
is just E to the T.
So, now applying integration by parts,
we're going to have UV
minus the integral of V, DU.
Well, U times V is T,
E to the T,
and we have minus the integral
from zero to three, V, DU,
which would be E to the T, DT,
and we still have plus
the integral from zero to three
of E to the T, DT.
You'll notice here these integrals are opposites,
and therefore, this sum is going to be zero,
so we just need to evaluate this
to find the value of the line integral.
So when T is equal to three,
we have three times E raised to the power of three
minus when T is zero,
we'd have zero times E to the zero,
which would just be zero times one or zero.
So the exact value of the line integral
is three times E raised to the power of three,
which is a decimal,
is approximately 60.2566.
I hope you found this helpful.